Muhammad Haris Rao
Let $\chi_P : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$ be the characteristic function of the primes. Recall the following functions \begin{align*} \pi(x) &= \sum_{n \le x} \chi_P(n) \\ \vartheta(x) &= \sum_{p \le x} \chi_P(n) \log{n} \\ \psi(x) &= \sum_{n \le x} \Lambda(n) \end{align*} where $\Lambda : \mathbb{Z}_{> 0} \longrightarrow \mathbb{C}$ is the von Mangoldt function defined by $\Lambda(n) = \log{p}$ if $n = p^m$ for some prime $p$ and $m \ge 1$, and $\Lambda(n) = 0$ otherwise. $\pi$ is called the prime counting function, and $\vartheta, \psi$ the first and second Chebychev functions respectively. The following assymptotic result is known as the prime number theorem:
Theorem (Prime Number Theorem): $\pi(x) \sim x / \log{x}$. That is, \begin{align*} \lim_{x \to \infty} \frac{\pi(x)}{x/ \log{x}} = 1 \end{align*}
Most proofs of the above fact actually work by proving an equivalent assymptotic result about the Chebychev functions rather than the result about $\pi$ directly. In particular, we have the following:
Theorem: The following are equivalent:
Proof. First, we show that the first two statements are equivalent. The key fact is using the Abel summation formula which tells us that \begin{align*} \vartheta(x) &= \sum_{n \le x} \chi_P(n) \log{n} = \log{x} \sum_{n \le x} \chi_P(n) - \int_1^x \frac{1}{t} \sum_{n \le t} \chi_P(n) \, dt = \pi(x) \log{x} - \int_1^x \frac{\pi(t)}{t} \, dt \\ \frac{\vartheta(x)}{x} &= \frac{\pi(x)}{x/\log{x}} - \frac{1}{x} \int_1^x \frac{\pi(t)}{t} \, dt \end{align*} Supposing that $\pi(x) \sim x / \log{x}$, there is $M > 0$ with $\pi(x) / (x / \log{x}) \le M$ and so \begin{align*} \frac{1}{x} \int_1^x \frac{\pi(t)}{t} \, dt &= \frac{1}{x} \int_1^{\sqrt{x}} \frac{\pi(t)}{t} \, dt + \frac{1}{x} \int_{\sqrt{x}}^x \frac{\pi(t)}{t} \, dt \\ &\le \frac{1}{x} \int_1^{\sqrt{x}} 1 \, dt + \frac{1}{x} \int_{\sqrt{x}}^x \frac{\pi(t)}{t/\log{t}} \frac{1}{\log{t}} \, dt \\ &\le \frac{\sqrt{x} - 1}{x} + \frac{M}{x} \int_{\sqrt{x}}^x \frac{1}{\log{t}} \, dt \\ &\le \frac{\sqrt{x} - 1}{x} + \frac{M}{x} \frac{x - \sqrt{x}}{\log{\sqrt{x}}} \\ \end{align*} Taking $x \to \infty$, everything vanishes so that given the prime number theorem, \begin{align*} \lim_{x \to \infty} \frac{\vartheta(x)}{x} = \lim_{x \to \infty} \frac{\pi(x)}{x/\log{x}} - \lim_{x \to \infty} \frac{1}{x} \int_1^x \frac{\pi(t)}{t} \, dt = \lim_{x \to \infty} \frac{\pi(x)}{x/\log{x}} = 1 \end{align*} which is (2). So (1) implies (2). Similarly, we have \begin{align*} \pi(x) &= 1 + \sum_{2 < n \le x} \chi_P(n) \\ &= 1 + \sum_{2 < n \le x} \frac{\chi_P(n)\log{n}}{\log{n}} \\ &= 1 + \frac{1}{\log{x}} \sum_{n \le x} \chi_P(n) \log{n} - \frac{1}{\log{2}} \sum_{n \le 2} \chi_P(n) \log{n} + \int_2^x \frac{1}{t \log^2{t}} \sum_{n \le t} \chi_P(n) \log{n} \, dt \\ &= \frac{\vartheta(x)}{\log{x}} + \int_2^x \frac{\vartheta(t)}{t \log^2{t}} \, dt \\ \frac{\pi(x)}{x/\log{x}} &= \frac{\vartheta(x)}{x} + \frac{\log{x}}{x} \int_2^x \frac{\vartheta(t)}{t \log^2{t}} \, dt \\ \end{align*} Assuming (2), we have $\vartheta(x) / x < M$ for some $M > 0$ and so by splitting the integral on $[2, x]$ into integrals on $[2, \sqrt{x}]$ and $[\sqrt{x}, x]$ we obtain the inequality \begin{align*} 0 \le \frac{\log{x}}{x} \int_2^x \frac{\vartheta(t)}{t \log^2{t}} \, dt \le \frac{M \log{x}}{x} \int_2^x \frac{1}{\log^2{t}} \, dt \le \frac{M \log{x}}{x} \left( \frac{\sqrt{x} - 2}{\log^2{2}} + \frac{x - \sqrt{x}}{\log^2{\sqrt{x}}} \right) \le \frac{M}{\log^2{2}} \frac{\log{x}}{\sqrt{x}} + \frac{4M}{\log{x}} \end{align*} Again taking $x \to \infty$, everything goes to 0. Assuming $\vartheta(x) \sim x$, then \begin{align*} \lim_{x \to \infty} \frac{\pi(x)}{x/\log{x}} = \lim_{x \to \infty} \frac{\vartheta(x)}{x} + \lim_{x \to \infty} \frac{\log{x}}{x} \int_2^x \frac{\vartheta(t)}{t \log^2{t}} \, dt = \lim_{x \to \infty} \frac{\vartheta(x)}{x} = 1 \end{align*} which is (1). So we have proven the equivalence between (1) and (2), and it is left to show that (2) and (3) are equivalent. We have for $x \ge 2$ that \begin{align*} \psi(x) &= \sum_{n \le x} \Lambda(n) = \sum_{k = 1}^\infty \sum_{p \le x^{1/k}} \log{p} = \sum_{k = 1}^\infty \vartheta(x^{1/k}) = \sum_{k \le \log_2{x}} \vartheta(x^{1/k}) = \vartheta(x) + \sum_{1 < k \le \log_2{x}} \vartheta(x^{1/k}) \end{align*} Hence, it follows that \begin{align*} 0 &\le \frac{\psi(x)}{x} - \frac{\vartheta(x)}{x} = \frac{1}{x} \sum_{1 < k \le \log_2{x}} \vartheta(x^{1/k}) \le \frac{1}{x} \sum_{1 < k \le \log_2{x}} \vartheta(x^{1/2}) \le \frac{\vartheta\left(\sqrt{x}\right) \log_2{x}}{x} \end{align*} It is not hard to see that $\vartheta\left( \sqrt{x} \right) \le \sqrt{x} \log{\sqrt{x}}$ so the last expression vanishes as $x \to \infty$. Then by the sandwich theorem, \begin{align*} \lim_{x \to \infty} \left( \frac{\psi(x)}{x} - \frac{\vartheta(x)}{x} \right) = 0 \end{align*} which proves the equivalence between (2) and (3). This completes the proof.
Next, define the function $\psi_1 : [1, \infty) \longrightarrow \mathbb{C}$ by \begin{align*} \psi_1 (x) &= \int_1^x \sum_{n \le t} \Lambda(n) \, dt = \int_1^x \psi(t) \, dt \end{align*} The assymptotics of $\psi$ can also be related to the assymptotics of $\psi_1$ as follows:
Proposition: If $\psi_1(x) \sim x^2 /2$, then $\psi(x) \sim x$.
Proof. Let $\alpha > 1$ be arbitrary. We have \begin{align*} \psi_1(\alpha x) - \psi_1(x) &= \int_x^{\alpha x} \sum_{n \le t} \Lambda(n) \, dt \le (\alpha x - x) \sum_{n \le \alpha x} \Lambda(n) = x \left( \alpha - 1 \right) \psi(\alpha x) \end{align*} Dividing through by $x^2/2$ yields \begin{align*} \alpha^2 \frac{\psi_1(\alpha x)}{(\alpha x)^2 / 2} - \frac{\psi_1(x)}{x^2/2} &\le \frac{2 \alpha \left( \alpha - 1 \right) \psi(\alpha x)}{\alpha x} \end{align*} The left hand side converges as $x \to \infty$ since $\psi_1(x) \sim x^2 / 2$ so the right side is bounded below. Thus, its inferior limit exists as \begin{align*} 2 \alpha (\alpha - 1) \liminf_{x \to \infty} \frac{\psi(\alpha x)}{\alpha x} &\ge \lim_{x \to \infty} \left( \alpha^2 \frac{\psi_1(\alpha x)}{(\alpha x)^2 / 2} - \frac{\psi_1(x)}{x^2/2} \right) = \alpha^2 - 1 \\ \liminf_{x \to \infty} \frac{\psi(\alpha x)}{\alpha x} \ge \frac{\alpha + 1}{2 \alpha} \end{align*} and this is true for all $\alpha > 1$. Then getting rid of the $\alpha$ in the inferior limit and taking $\alpha \to 1^+$ on the right and gives us \begin{align*} \liminf_{x \to \infty} \frac{\psi(x)}{x} \ge 1 \end{align*} Now, suppose that $0 < \alpha < 1$. This time we have \begin{align*} \psi_1(x) - \psi_1(\alpha x) &= \int_{\alpha x}^x \sum_{n \le t} \Lambda(n) \, dt \ge (x - \alpha x) \sum_{n \le \alpha x} \Lambda(n) = x(1 - \alpha) \psi(\alpha x) \\ \frac{\psi_1(x)}{x^2 / 2} - \alpha^2 \frac{\psi_1(\alpha x)}{(\alpha x)^2 / 2} &\ge \frac{2 \alpha (1 - \alpha) \psi(\alpha x)}{\alpha x} \end{align*} Taking $x \to \infty$, gets us \begin{align*} 2 \alpha (1 - \alpha)\limsup_{x \to \infty} \frac{\psi(\alpha x)}{\alpha x} &\le \lim_{x \to \infty} \left( \frac{\psi_1(x)}{x^2 / 2} - \alpha^2 \frac{\psi_1(\alpha x)}{(\alpha x)^2 / 2} \right) = 1 - \alpha^2 \\ \limsup_{x \to \infty} \frac{\psi(\alpha x)}{\alpha x} &\le \frac{1 + \alpha}{2 \alpha} \end{align*} Again, ridding of the $\alpha$ inside the limit superior and taking $\alpha \to 1^-$ gets \begin{align*} \limsup_{x \to \infty} \frac{\psi(x)}{x} \le 1 \end{align*} Hence, we have shown overall that \begin{align*} \limsup_{x \to \infty} \frac{\psi(x)}{x} \le 1 \le \liminf_{x \to \infty} \frac{\psi(x)}{x} \end{align*} This can only mean that the inequalities are in fact equalities since the superior limit is always at least as much as the inferior. This proves $\psi(x) \sim x$ which is what we claimed.
The consequence of all this is that the prime number theorem is implied by the limit \begin{align*} \lim_{x \to \infty} \frac{1}{x^2} \int_1^x \sum_{n \le t} \Lambda(n) \, dt = \frac{1}{2} \end{align*}